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THE METHOD OF SECTIONS Today’s Objectives: Students will be able to determine: Forces in truss members using the method of sections In-Class Activities: • Check Homework, if any • Reading Quiz • Applications • Method of Sections • Concept Quiz • Group Problem Solving • Attention Quiz Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved READING QUIZ In the method of sections, generally a “cut” passes through no more than _ members in which the forces are unknown A) B) C) D) If a simple truss member carries a tensile force of T along its length, then the internal force in the member is A) Tensile with magnitude of T/2 B) Compressive with magnitude of T/2 C) Compressive with magnitude of T D) Tensile with magnitude of T Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved APPLICATIONS Long trusses are often used to construct large cranes and large electrical transmission towers The method of joints requires that many joints be analyzed before we can determine the forces in the middle of a large truss So another method to determine those forces is helpful Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved THE METHOD OF SECTIONS In the method of sections, a truss is divided into two parts by taking an imaginary “cut” (shown here as a-a) through the truss Since truss members are subjected to only tensile or compressive forces along their length, the internal forces at the cut members also will be either tensile or compressive, with the same magnitude as the forces at the joint This result is based on the equilibrium principle and Newton’s third law Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved STEPS FOR ANALYSIS Decide how you need to “cut” the truss This is based on: a) where you need to determine forces, and, b) where the total number of unknowns does not exceed three (in general) Decide which side of the cut truss will be easier to work with (goal is to minimize the number of external reactions) If required, determine any necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved STEPS FOR ANALYSIS (continued) Draw the FBD of the selected part of the cut truss You need to indicate the unknown forces at the cut members Initially, you may assume all the members are in tension, as done when using the method of joints Upon solving, if the answer is positive, the member is in tension, as per the assumption If the answer is negative, the member is in compression (Please note that you can assume forces to be either tension or compression by inspection as was done in the figures above.) Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved STEPS FOR ANALYSIS (continued) Apply the scalar equations of equilibrium (E-of-E) to the selected cut section of the truss to solve for the unknown member forces Please note, in most cases it is possible to write one equation to solve for one unknown directly So look for it and take advantage of such a shortcut! Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE Given: Loads as shown on the truss Find: The force in members KJ, KD, and CD Plan: a) Take a cut through members KJ, KD and CD b) Work with the left piece of the cut sections Why? c) Determine the support reactions at A What are they? d) Apply the E-of-E to find the forces in KJ, KD and CD Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE (continued) AX AY GY Analyzing the entire truss for the reactions at A, we get FX = AX = A moment equation about G to find AY results in: MG = AY (12) – 20 (10) – 30 (8) – 40 (6) = 0; Statics, Fourteenth Edition R.C Hibbeler AY = 56.7 kN Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE (continued) FKJ FKD FCD 56.7 kN Now take moments about point D Why this? + MD = – 56.7 (6) + 20 (4) + 30 (2) – FKJ (3) = FKJ = − 66.7 kN or 66.7 kN ( C ) Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved EXAMPLE (continued) FKJ FKD FCD 56.7 kN Now use the x and y-directions equations of equilibrium ↑ + FY = 56.7 – 20 – 30 – (3/13) FKD = 0; FKD = 8.05 kN (T) → + FX = (– 66.7) + (2/13) ( 8.05 ) + FCD = 0; FCD = 62.2 kN (T) Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CONCEPT QUIZ Can you determine the force in member ED by making the cut at section a-a? Explain your answer A) No, there are four unknowns B) Yes, using MD = C) Yes, using ME = D) Yes, using MB = Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved CONCEPT QUIZ (continued) If you know FED, how will you determine FEB? A) By taking section b-b and using ME = B) By taking section b-b, and using FX = and FY = C) By taking section a-a and using MB = D) By taking section a-a and using MD = Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING Given: Loads as shown on the truss Find: The forces in members ED, EH, and GH Plan: a) Take the cut through members ED, EH, and GH b) Analyze the left section Determine the support reactions at F Why? c) Draw the FBD of the left section d) Apply the equations of equilibrium (if possible, try to it so that every equation yields an answer to one unknown Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING (continued) 1) Determine the support reactions at F by drawing the FBD of the entire truss Fy Ax Ay + MA = – Fy (4) + 40 (2) + 30 (3) + 40 (1.5) = 0; Fy = 57.5 kN Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved GROUP PROBLEM SOLVING (continued) 2) Analyze the left section + ME = – 57.5 (2) + FGH (1.5) = 0; FGH = 76.7 kN (T) FED 1.5 m ↑ + Fy = 57.5 – 40 – FEH (3/5)= 0; FEH = 29.2 kN (T) FGH Fy= 57.5 kN + MH = – 57.5 (4) + 40 (2) – FED (1.5) = 0; FED = -100 kN = 100 kN (C) Statics, Fourteenth Edition R.C Hibbeler FEH Copyright ©2016 by Pearson Education, Inc All rights reserved ATTENTION QUIZ As shown, a cut is made through members GH, BG and BC to determine the forces in them Which section will you choose for analysis and why? A) Right, fewer calculations B) Left, fewer calculations C) Either right or left, same amount of work D) None of the above, too many unknowns Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved ATTENTION QUIZ When determining the force in member HG in the previous question, which one equation of equilibrium is the best one to use? A) MH = B) MG = C) MB = D) MC = Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved End of the Lecture Let Learning Continue Statics, Fourteenth Edition R.C Hibbeler Copyright ©2016 by Pearson Education, Inc All rights reserved ... Edition R. C Hibbeler Copyright 20 16 by Pearson Education, Inc All rights reserved APPLICATIONS Long trusses are often used to construct large cranes and large electrical transmission towers The... necessary support reactions by drawing the FBD of the entire truss and applying the E-of-E Statics, Fourteenth Edition R. C Hibbeler Copyright 20 16 by Pearson Education, Inc All rights reserved... (3) = FKJ = − 66 .7 kN or 66 .7 kN ( C ) Statics, Fourteenth Edition R. C Hibbeler Copyright 20 16 by Pearson Education, Inc All rights reserved EXAMPLE (continued) FKJ FKD FCD 56. 7 kN Now use
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